The area of a rectangle is 21 yd.² and the length of a rectangle is 1 yard or less than twice the width find the dimensions of the rectangle

length = 6 yards

width = 3.5 yards

Step-by-step explanation:

the length of a rectangle is 1 yard less than twice the width

LEt 'w' be the width of the rectangle

and 'l' be the length of the rectangle

length l = 2 times the width -1

so l = 2w -1

Area of the rectangle = length * width

Area = l * w

we know length is 2w-1  and given area = 21

21 = (2w-1) * w

subtract 21 on both sides

now solve for 'w'

0 = 2w(w+3)-7(w+3)

0= (2w-7)(w+3)

2w-7 = 0                    and w+3 =0

w = 7/2                            w=-3

width cannot be negative so we ignore w=-3

width w= 3.5

length = 2w-1= 2(3.5) - 1= 7-1=6

length = 6 yards

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %